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3 years ago

11

The attractive electrostatic force between the point charges 6.14×10−6 c and q has a magnitude of 0.845 n when the separation be

tween the charges is 9.26 m . part a find the sign and magnitude of the charge q.

1 answer:

Nataly [62]3 years ago

3 0

Coulomb's law states that F = k(q1)(q2)/r^2, where k = 9 x 10^9 N-m^2/C^2
Substituting the given values:

0.845 N = (9x10^9 N-m^2/C^2)(6.14x10^-6 C)(q) / (9.26^2)
q = 0.00131 = 1.31 x 10^-3 C

Since the force is attractive, the point charges must be of opposite signs, and q is negatively charged (-1.31 x 10^-3 C).

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Melt.

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A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

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$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

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3 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

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3 years ago

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Answer:

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The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

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where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

$v=\frac{c}{n}$

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1 ,n_2$ are the index of refraction of the 1st and 2nd medium

$\theta_1,\theta_2$ are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

So we see that if the index of refraction of the second medium is lower ( $n_2$ ), then the ratio $\frac{n_1}{n_2}$ is larger than 1, so the angle of refraction is larger than the angle of incidence:

$\theta_2>\theta_1$

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

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3 years ago